Kohn-Sham Equations

Using what we have just learned about taking derivatives of real functions of complex variables and including the normality constraint of each orbital $\psi_i(\vec x)$ with a separate Lagrange multiplier $\lambda_i$, the condition for the constrained minimization of (12) is

\begin{eqnarray*}
0 & = & \frac{\delta}{\delta \psi_i^*(\vec x)}\left( T_{el} + ...
..._i
\lambda_i \int \psi_i^*(\vec x) \psi_i(\vec x) \, dV \right).
\end{eqnarray*}


We now take this derivative term by term.

The kinetic energy (10) has only one $\psi_i^*(\vec x)$ term in it, so the derivative is just what this term multiplies,

\begin{displaymath}
\frac{\delta}{\delta \psi_i^*(\vec x)}\left( T_{el} \right) =
-f_i\, \frac{\hbar^2}{2m} \psi_i(\vec x).
\end{displaymath}

For the electron-nuclear potential energy (3), the only term which depends on $\psi_i^*(\vec x)$ is the charge density, where $\psi_i^*(\vec x)$ multiplies $f_i\, \psi_i(\vec x)$. The nuclear potential $V_{nuc}(\vec x)$ is unchanged as $\psi_i^*(\vec x)$ varies, so the final term is just

\begin{displaymath}
\frac{\delta}{\delta \psi_i^*(\vec x)}\left( U_{el-nuc} \right) =
f_i \, V_{nuc}(\vec x) \psi_i(\vec x).
\end{displaymath}

The electron-electron energy has a very similar structure. The only difference is that, here, when we change $\psi_i^*(\vec x)$, the potential function $\phi(\vec x)$ also changes because it depends on $n(\vec
x)$. The net effect of the change in $\phi(\vec x)$ is the same as that of the direct change in $n(\vec
x)$. To see this we note that Poisson's equation $\nabla^2 \phi = - 4 \pi [k_c] e^2 n$ implies also that $\nabla^2
(\delta\phi) = - 4 \pi [k_c] e^2 (\delta n)$. Thus,

\begin{displaymath}
\int (\delta \phi) n \, dV = \int (\delta \phi) \frac{\nabla...
...4 \pi [k_c] e^2}\right) \phi\,dV
= \int (\delta n) \phi \, dV,
\end{displaymath}

where we have moved the $\nabla^2$ from acting on $\phi$ to acting on $\delta\phi$ by integrating by parts twice. Since both terms are equal, we can take just twice the $(1/2) \int \phi \, \delta n\,dV$ term,

\begin{displaymath}
\frac{\delta}{\delta \psi_i^*(\vec x)}\left( U_{el-el} \right) =
f_i \, \phi(\vec x) \psi_i(\vec x).
\end{displaymath}

For the exchange-correlation term, we have already derived in (14) that $\delta E_{xc}/\delta n(\vec x) = f'_{xc}(n(\vec
x))$. By the chain rule we just need to multiply this by $\delta
n(\vec x)/\delta \psi_i^*(\vec x)=f_i\, \psi_i(\vec x)$ for the result

\begin{displaymath}
\frac{\delta}{\delta \psi_i^*(\vec x)}\left( E_{xc} \right) = f_i \,
f'_{xc}(n(\vec x)) \psi_i(\vec x).
\end{displaymath}

Fortunately, $U_{nuc-nuc}$ depends only on the nuclear positions and does not change with $\psi_i^*(\vec x)$, so

\begin{displaymath}
\frac{\delta}{\delta \psi_i^*(\vec x)}\left( U_{nuc-nuc} \right) = 0.
\end{displaymath}

And, finally, $\psi_i^*(\vec x)$ only appears once in the constraint term, making the derivative,

\begin{displaymath}
\frac{\delta}{\delta \psi_i^*(\vec x)}\left( - \sum_i
\lambd...
... x) \psi_i(\vec x) \, dV \right) =
- \lambda_i \psi_i(\vec x).
\end{displaymath}

Summing all of these contributions, setting the resulting equation to zero, moving the `` $\lambda_i \psi_i(\vec x)$'' term to the right-hand side, and dividing through by $f_i$, we get the final result,

\begin{displaymath}
-\frac{\hbar^2}{2m} \nabla^2 \psi_i(\vec x) + \left[V_{nuc}(...
...ight] \psi_i(\vec x) =
\frac{\lambda_i}{f_i} \psi_i(\vec x).
\end{displaymath}

Fortunately for us, this is in the form of a very well-known equation for which there are standard techniques. This is in the form of the standard Schrödinger equation,
\begin{displaymath}
-\frac{\hbar^2}{2m} \nabla^2 \psi_i(\vec x) + V(\vec x) \psi_i(\vec x) =
\epsilon_i \psi_i(\vec x),
\end{displaymath} (16)

where we define the potential term as
\begin{displaymath}
V(\vec x) \equiv V_{nuc}(\vec x) + \phi(\vec x) + f'_{xc}(n(\vec x)),
\end{displaymath} (17)

and we define $\epsilon_i \equiv \lambda_i/f_i$. We interpret the potential $V(\vec x)$ as just the sum of the nuclear potential, the electrostatic potential $\phi(\vec x)$ created by the electrons, and an extra, ``exchange-correlation'' potential correction, $V_{xc}(\vec
x)\equiv f'_{xc}(n(\vec x))$. Since the Lagrange-multipliers are unknown constants at the start, we may as well think in terms of the $\epsilon_i \equiv \lambda_i/f_i$ instead, which have the interpretation of the Schrödinger energies for each orbital.

Tomas Arias 2004-01-26